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Triplets & Triangles Author
Posts: 63
Location: United Kingdom | As Thomas has observed, I simply tried to solve the Sudokus in order. This was much less stressful than continually shuffling bits of paper. In my case, it took 10 minutes for the first puzzle to be printed (as the pdf file was very slow to open and printing was also incredibly slow). As a result, the unusual scoring system really penalised me. My strategy was also flawed by the fact that I got really stuck on the Distance Sudoku. I would prefer a conventional scoring system.
Some beautiful puzzles, including those which I solved afterwards. Particularly liked Number 5 Still Alive, Group Sum and Odd-Even-Big-Small. |
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Location: India | David McNeill - 2011-07-18 10:00 PM
(as the pdf file was very slow to open and printing was also incredibly slow).
Really surprised to hear that, because the pdf file size was really small (~100 KB), and all grids were vector based images (except some part of odd-even-big-small).
Did anybody else face the problem?
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Posts: 89
Location: India | I would second david's point on the fact that puzzles should be sequenced as per points,
but i loved the time based scoring system, knowing your strengths and selecting the right strategy should be rewarded |
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Posts: 1
| can any body help out for how to start the "crossnumber sudoku and number 5 still alive"?
Edit : Removed a link not related to sudoku / puzzles
Edited by debmohanty 2011-12-08 9:15 AM
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WPMM Author
Posts: 66
Location: Hungary | atc121212 - 2011-12-08 9:10 AM
can any body help out for how to start the "crossnumber sudoku and number 5 still alive"?
Edit : Removed a link not related to sudoku / puzzles
Crossnumber: There are just 4 3-long and 4-long numbers. Not too hard to find their places. Four instance in the 1st row the 3-long number can be just one from the list.
After the 3-long number in the last row also unique. You can put the other two 3-long number easily.
Now you can find the places of 4-long numbers.
Number 5: A bit harder (it is 78 points)
Four instance in the left-middle box there are 3 cages. The sum 5+5+15 or 5+15+15 But the second isnt good, because the total of remained cells would be 10, and a 6 is given. So it is 1+3+6, but R3C1=3
Hence the correct sum 5+5+15 so this 3 cages contain (2+3) (1+4) (7+8) and the 2 remained cells (5,9)
You can consider the other side as well, the 2 remained cells (1,5)
You could write down numbers, but these are important informations.
If you look at the 5th column you will see 4 cages. The total of this cages 40 or 35 or 30 etc. So 40. And R5C5=5. The number pairs in this column (14) (23) (69) (78) and just one cell can be 9. (R3C5) Now you can find the 9 and 6 in the right-middle box. (5,9) in the left-middle box and (1,5) in the right-middle box. Moreover all 9s unique.
I hope this will be enough.
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